Question

How do you find the point of intersection of\[f(x) = x\] and\[g(x) = - x - 2?\]

Answer 1

Hint:The given question describes the operation of addition/ subtraction/ multiplication/ division. We need to know the condition which is used to find the intersection points of two lines. In this question, we have to find the value\[x\]by using a suitable formula. Also, we need to know the multiplication between a positive number and a negative number.

Complete step by step solution:
To solve the given question, we have to find the intersection point of \[f(x) = x\]and\[g(x) = - x - 2\].

The condition to find the intersection point between the given equations is,
\[f(x) = g(x) \to \left( 1 \right)\]

We know that,\[f(x) = x\] \[g(x) = - x - 2\]

Let’s substitute the above values in the equation\[\left( 1 \right)\] we get,

\[f(x) = g(x)\]
\[x = - x - 2\]

To make easy calculations we have to add\[x\] LHS and RHS to the equation. So, we get
\[x + x = - x - 2 + x\]

On the right-hand side of the equation, we have\[ - x\]and\[x\], both are cancelled by each other.

So, we get

\[x + x = 2 \to \left( 2 \right)\]
We know that, \[x + x\]
\[ \downarrow \]
\[2x\]

So, the equation\[\left( 2 \right)\]becomes
\[2x = - 2\]

Let’s move the term\[2\]from the left side to right side of the equation in the\[2x\]form. So, we get

\[x = \dfrac{{ - 2}}{2}\]
\[x = - 1\]

So, the final answer is,
\[x = - 1\]Is the intersection point of,\[f(x) = x\]and \[g(x) = - x - 2\].

Note: The intersection point is the point at where the given two line values are equal\[(i.ef(x) = g(x))\].

To make an easy calculation we can add/ subtract/ multiply/ divide any term with the equation into both sides. We shouldn’t multiply zero with the equation, otherwise, we didn’t get the final answer. When multiplying the different sign numbers we would remind the following things,

\[\left( - \right) \times \left( + \right) = \left( - \right)\]

\[\left( - \right) \times \left( - \right) = \left( + \right)\]

\[\left( + \right) \times \left( + \right) = \left( + \right)\]